The drawing shows a block (m = 1.5 kg) and a spring (k = 300 N/m) on a frictionless incline. The spring is compressed by x0 = 0.34 m relative to its unstrained position at x = 0 m and then released. What is the speed of the block when the spring is still compressed by xf = 0.14 m?
Well, compute the spring energy at .34m,and at .14m. The difference is the amount of energy the spring has released. This energy goes to KE and change of GPE. You did not mention the slant angle, but GPE is gained (mgh)
spring energy released=KE+ mgh
of course spring energy= 1/2 k x^2, here the amount released is 1/2 k(.34^2-.14^2)
To find the speed of the block when the spring is compressed by xf = 0.14 m, we can use the principle of conservation of mechanical energy.
The potential energy stored in the compressed spring is given by the equation:
PE = (1/2)k(xf^2 - x0^2)
where k is the spring constant, xf is the final compression distance, and x0 is the initial compression distance.
The potential energy is converted into kinetic energy as the block moves down the incline. The kinetic energy (KE) of the block is given by the equation:
KE = (1/2)mv^2
where m is the mass of the block and v is its velocity.
According to the principle of conservation of mechanical energy, the total mechanical energy (PE + KE) remains constant throughout the motion of the block.
Therefore, we can equate the initial potential energy of the compressed spring to the final kinetic energy of the block:
(1/2)k(xf^2 - x0^2) = (1/2)mv^2
Plugging in the given values:
k = 300 N/m
xf = 0.14 m
x0 = 0.34 m
m = 1.5 kg
we can solve for v using the above equation.
(1/2)(300)(0.14^2 - 0.34^2) = (1/2)(1.5)v^2
Simplifying the equation:
(1/2)(300)(0.0196 - 0.1156) = (1/2)(1.5)v^2
(1/2)(300)(-0.096) = (1/2)(1.5)v^2
-14.4 = (1/2)(1.5)v^2
Dividing both sides by (1/2)(1.5):
v^2 = -9.6
Since speed cannot be negative, we can ignore the negative sign:
v^2 = 9.6
Taking the square root of both sides:
v ≈ 3.1 m/s
Therefore, the speed of the block when the spring is still compressed by xf = 0.14 m is approximately 3.1 m/s.